MHT CET 2017 :: Physics :: General questions

• Physics - General questions

A simple pendulum of length 'L' has mass 'M' and it oscillates freely with amplitude 'A'. At the extreme position, its potential energy is 
(g= acceleration due to gravity)

Answer:

Explanation:

Potential energy of a simple pendulum is given as,

$=\frac{1}{2}M\omega^{2}A^{2}=\frac{1}{2}M.\frac{g}{L}.A^{2}$    $\left(\because  \omega = \sqrt{\frac{g}{L}}\right)$

The fundamental frequency of an air column in a pipe closed at one end is 100 Hz. If the same pipe is open at both ends, the frequencies produced in Hz are 

Answer:

Explanation:

 For a closed pipe fundamental frequency, 

$v_{1}=\frac{v}{4L}=100 Hz$

For an open pipe fundamental frequency,

$v_{1}=\frac{v}{2L}=200 Hz$

 In a pipe open at both the ends , all multiples of the fundamental are produced

The depth 'd' at which the value of acceleration  due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is 

(R= radius of earth)

Answer:

Explanation:

 Acceleration due to gravity at depth d is given as, 

     g'= $g\left(1-\frac{d}{R}\right)$   ...........(i)

 Given,  $g'=\frac{g}{n}$

 Substituting the value of 'g' in Eq.(i) , we  get

$\frac{g}{n}=g\left(1-\frac{d}{R}\right)$

 $\Rightarrow $    $\frac{1}{n}=1-\frac{d}{R}$

$\therefore$  $ \frac{d}{R}=1-\frac{1}{n}=\frac{n-1}{n}$

 $d=R\left(\frac{n-1}{n}\right)$

The ratio of binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of the same mass  revolving around the earth at a height h above the earth's surface is (R= radius of the earth)

Answer:

Explanation:

 Binding energy on the surface of the earth is 

$E_{1}=\frac{GMm}{R}$         .............(i)

 Binding energy of revolving  satellite  at a height h is 

$E_{2}=\frac{GMm}{2(R+h)}$......(ii)

  From Eqs. (i) and (ii) , we get

$\therefore$     $\frac{E_{1}}{E_{2}}=\frac{2(R+h)}{R}$

A big water drop is formed by the combination of 'n' small water drops of equal radii. The ratio of the surface  energy of 'n'  drops to the surface energy of big drop is 

Answer:

Explanation:

 Volume of big drop = n(Volume of small drop)

 $\Rightarrow$     $\frac{4}{3}\pi R^{3}=n.\frac{4}{3}\pi r^{3}$

 where, R= radius of big water drop

and r = radius of small water drop

$\Rightarrow$  $R^{3}  =n r^{3}$

$\Rightarrow $     $R= n^{1/3}.r$  ........(i)

surface energy of n drops,

 $E_{2}= n \times 4 \pi r^{2}\times T$

Surface energy of big drop ,

  $E_{1}=  4 \pi R^{2}T$

$\therefore$     $\frac{E_{2}}{E_{1}}=  \frac{nr^{2}}{R^{2}}= \frac{nr^{2}}{(n^{1/3}.r)^{2}}$

  $= \frac{nr^{2}}{n^{2/3}.r^{2}}= n^{1/3}$     $  [\because $ from eq.(i)]

 $E_{2}:E_{1}=\sqrt[3]{n}:1$

• Physics - General questions